Question

If $E^{\circ}(\text{Cu}^{2+}/\text{Cu})=+0.34\text{ V}$. What is potential for $\text{Cu}(s)\rightarrow \text{Cu}^{2+}(aq)(0.1\text{M})+2e^{-}$ at 298 K?

• A. +0.3696 V
• B. -0.3696 V
• C. +0.3104 V
• D. -0.3104 V

Hint:

Read the requested reaction carefully! If it shows electrons on the right side ($X \rightarrow X^+ + e^-$), it's asking for the Oxidation Potential. You must flip the sign of the standard reduction potential before plugging it into Nernst!

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The Nernst equation is used to calculate the electrode potential under non-standard conditions. Note that the question asks for the oxidation potential ($Cu \to Cu^{2+}$).

Step 2: Formula Application:
First, find Reduction Potential ($E_{red}$): $E_{red} = E^{\circ} - \frac{0.059}{n} \log \frac{1}{[Cu^{2+}]}$ Then, Oxidation Potential ($E_{ox}$) = $-E_{red}$.

Step 3: Explanation:
$E_{red} = 0.34 - \frac{0.059}{2} \log \frac{1}{0.1}$ $E_{red} = 0.34 - 0.0295 \log(10) = 0.34 - 0.0295 = 0.3105$ V. Since we need the oxidation potential for $Cu \to Cu^{2+}$, $E_{ox} = -0.3105$ V.

Step 4: Final Answer:
The potential is -0.3104 V.